Respuesta :
Given:
μ = 200 lb, the mean
σ = 25, the standard deviation
For the random variable x = 250 lb, the z-score is
z = (x-μ)/σ =(250 - 200)/25 = 2
From standard tables for the normal distribution, obtain
P(x < 250) = 0.977
Answer: 0.977
μ = 200 lb, the mean
σ = 25, the standard deviation
For the random variable x = 250 lb, the z-score is
z = (x-μ)/σ =(250 - 200)/25 = 2
From standard tables for the normal distribution, obtain
P(x < 250) = 0.977
Answer: 0.977
Using the normal distribution, it is found that there is a 0.9772 = 97.72% probability of a player weighing less than 250 pounds.
Normal Probability Distribution
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
In this problem:
- The mean is of [tex]\mu = 200[/tex].
- The standard deviation is of [tex]\sigma = 25[/tex].
The probability of a player weighing less than 250 pounds is the p-value of Z when X = 250, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{250 - 200}{25}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a p-value of 0.9772.
0.9772 = 97.72% probability of a player weighing less than 250 pounds.
More can be learned about the normal distribution at https://brainly.com/question/24663213
